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Physics

ACT

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1

CHAPTER OUTLINE 1.1 Standards of Length, Mass, and Time 1.2 Matter and Model-Building 1.3 Density and Atomic Mass 1.4 Dimensional Analysis 1.5 Conversion of Units 1.6 Estimates and Order-of- Magnitude Calculations 1.7 Significant Figures

Physics and Measurement

ANSWERS TO QUESTIONS

Q1.1 Atomic clocks are based on electromagnetic waves which atoms emit. Also, pulsars are highly regular astronomical clocks.

Q1.2 Density varies with temperature and pressure. It would be necessary to measure both mass and volume very accurately in order to use the density of water as a standard.

Q1.3 People have different size hands. Defining the unit precisely would be cumbersome.

Q1.4 (a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms

Q1.5 (b) and (d). You cannot add or subtract quantities of different dimension.

Q1.6 A dimensionally correct equation need not be true. Example: 1 chimpanzee = 2 chimpanzee is dimensionally correct. If an equation is not dimensionally correct, it cannot be correct.

Q1.7 If I were a runner, I might walk or run 10 1 miles per day. Since I am a college professor, I walk about 10 0 miles per day. I drive about 40 miles per day on workdays and up to 200 miles per day on vacation.

Q1.8 On February 7, 2001, I am 55 years and 39 days old.

55 yr 365 25 1 .yr^ d 39 d 20 128 d 86 400 1 d s 1 74. 10 9 s ~ 109 s

F HG^

I KJ^

• = FHG IKJ = ×.

Many college students are just approaching 1 Gs.

Q1.9 Zero digits. An order-of-magnitude calculation is accurate only within a factor of 10.

Q1.10 The mass of the forty-six chapter textbook is on the order of 10 0 kg.

Q1.11 With one datum known to one significant digit, we have 80 million yr + 24 yr = 80 million yr.

Chapter 1 3

P1.6 For either sphere the volume is V = 4 r 3

π 3 and the mass is m = ρ V =ρ 4 π r 3

(^3). We divide this equation

for the larger sphere by the same equation for the smaller:

m m

r r

r s (^) s rs

A (^) = ρ π A^ = A = ρ π

3 3

3

Then r (^) A = rs^3 5 = 4 50. cm a1 71. f = 7 69. cm.

P1.7 Use 1 u = 1 66. × 10 −^24 g.

(a) For He, m 0 = 4 00 × 6 64 1024

F HG^

I KJ^

. u 1.66^10 g =. × − 1 u

g

• .

(b) For Fe, m 0 = 55 9 × 9 29 1023

F HG^

I KJ^

. u 1.66^10 g =. × − 1 u

g

• .

(c) For Pb, m 0

(^24 ) = 207 1 66^ ×^10 3 44 10

F HG^

I KJ^

= ×

− (^) − u g 1 u

. (^). g.

*P1.8 (a) The mass of any sample is the number of atoms in the sample times the mass m 0 of one atom: m = Nm 0. The first assertion is that the mass of one aluminum atom is

m 0 = 27 0. u = 27 0. u × 1 66. × 10 −^27 kg 1 u = 4 48. × 10 −^26 kg. Then the mass of 6 02. × 10 23 atoms is m = Nm 0 = 6 02. × 10 23 × 4 48. × 10 −^26 kg = 0 027 0. kg =27 0. g. Thus the first assertion implies the second. Reasoning in reverse, the second assertion can be written m = Nm 0.

0 027 0. kg = 6 02. × 10 23 m 0 , so m 0 0 027 23 26 6 02 10

×

. = × −

kg (^). kg ,

in agreement with the first assertion.

(b) The general equation m = Nm 0 applied to one mole of any substance gives M g = NM u, where M is the numerical value of the atomic mass. It divides out exactly for all substances, giving 1 000 000 0. × 10 −^3 kg = N 1 660 540 2. × 10 −^27 kg. With eight-digit data, we can be quite sure of the result to seven digits. For one mole the number of atoms is

N = F HG^

I KJ^

1 − + = ×

(c) The atomic mass of hydrogen is 1.008 0 u and that of oxygen is 15.999 u. The mass of one molecule of H O 2 is 2 1 008 0b. (^) g+ 15 999. u =18 0. u. Then the molar mass is 18 0. g.

(d) For CO 2 we have 12 011. g + 2 15 999b. g (^) g = 44 0. g as the mass of one mole.

4 Physics and Measurement

P1.9 Mass of gold abraded: ∆ m = − = =

F HG^

I KJ^

3 80. g 3 35. g 0 45. g 0 45. g 1 kg = 4 5. × 10 −^4 10 g

b g 3 kg.

Each atom has mass m 0

(^2725) 197 197 1 66^10 1

= = × 3 27 10

F HG^

I KJ^

= ×

− − u u kg u

. (^). kg.

Now, ∆ m = ∆ N m 0 , and the number of atoms missing is

N

m m

= = ×

×

= ×

− 0 −

4 25

kg. kg

atoms.

The rate of loss is

∆ ∆ ∆ ∆

N

t N t

= ×^ FHG IKJFHG IKJFHG IKJFHG IKJ

= ×

21

11

atoms yr

yr 365.25 d

d 24 h

h 60 min

min 60 s

atoms s

P1.10 (a) m = ρ L^3 = 3 × −^6 = × −^ = × −

(^3 16 ) e7 86.^ g cm^ je5 00.^10 cm^ j 9 83.^10 g^ 9 83.^10 kg

(b) N m m

= = ×

×

= ×

− 0 −

19 27

kg. u kg 1 u

atoms e j

P1.11 (a) The cross-sectional area is

A = + = × −

m m m m m 2

a (^) fa (^) f a (^) fa (^) f .

The volume of the beam is

V = AL = (^) e 6 40. × 10 −^3 m (^2) ja1 50. m f = 9 60. × 10 −^3 m^3.

Thus, its mass is

m = ρ V = (^) e7 56. × 10 3 kg / m (^3) je9 60. × 10 −^3 m (^3) j = 72 6. kg.

FIG. P1.

(b) The mass of one typical atom is m 0

(^2726) 55 9 1 66^10 1

= × 9 28 10

F HG^

I KJ^

= ×

− −

. u.^ kg. u

a f kg. Now

m = Nm 0 and the number of atoms is N m m

×

− =^ ×

0 26

kg. kg

atoms.

6 Physics and Measurement P1.15 (a) This is incorrect since the units of ax are m 2 s^2 , while the units of v are m s.

(b) This is correct since the units of y are m, and cos a f kx is dimensionless if k is in m−^1.

*P1.16 (a) a

F

∝ (^) m

∑ or a k F

= (^) m

∑ represents the proportionality of acceleration to resultant force and

the inverse proportionality of acceleration to mass. If k has no dimensions, we have

a k F m

= , L

T

1 F

2 M

= , F M L

T 2

= ⋅^.

(b) In units, M L T

kg m (^2) s 2

⋅ (^) = ⋅ (^) , so 1 newton = 1 kg m s⋅ 2.

P1.17 Inserting the proper units for everything except G ,

kg m s

kg (^2) m

L NM^

O QP^

G^2

Multiply both sides by m 2 and divide by kg 2 ; the units of G are m kg s

3 ⋅ 2.

Section 1.5 Conversion of Units

*P1.18 Each of the four walls has area a8 00. ft fa12 0. ft f= 96 0. ft 2. Together, they have area

4 96 0 (^2) 3 28^1 35 7

2 e.^ ft^ jFHG^. mft IKJ^ =^. m^2.

P1.19 Apply the following conversion factors:

1 in = 2 54. cm, 1 d = 86 400s, 100 cm = 1 m, and 10 9 nm = 1 m

2 9 in day

cm in m cm nm m 86 400 s day

FHG IKJ = nm s

b ge je j .

This means the proteins are assembled at a rate of many layers of atoms each second!

*P1.20 8 50 8 50 0 025 4^ 1 39 10

3

. in. in.^ m.^4 1 in

(^3) = 3 FHG IKJ = × − m 3

Chapter 1 7

P1.21 Conceptualize : We must calculate the area and convert units. Since a meter is about 3 feet, we should expect the area to be about A ≈ (^) a 30 m (^) fa 50 m (^) f =1 500m 2. Categorize : We model the lot as a perfect rectangle to use Area = Length × Width. Use the conversion: 1 m = 3.281 ft.

Analyze : A = LW = 100 FHG 1 IKJ FHG IKJ = × 3 281

ft m 1 39 10 3 ft

ft m ft

a f a f = 1 390 m 2 m^2

..

Finalize : Our calculated result agrees reasonably well with our initial estimate and has the proper units of m 2. Unit conversion is a common technique that is applied to many problems.

P1.22 (a) V = a40.0 m fa20.0 m fa 12.0 m f = 9 60. × 10 3 m 3

V = 9 60. × 10 3 m 3 b3.28 ft 1 m g 3 = 3 39. × 105 ft^3

(b) The mass of the air is

m = ρair V = (^) e1 20. kg m (^3) je9.60 × 10 3 m (^3) j = 1 15. × 104 kg.

The student must look up weight in the index to find

F (^) g = mg = (^) e1.15 × 10 4 kg (^) je9.80 m s 2 j = 1.13 × 10 5 N.

Converting to pounds,

Fg = (^) e1 13. × 10 5 N (^) jb1 lb 4.45 N g = 2 54. × 104 lb.

P1.23 (a) Seven minutes is 420 seconds, so the rate is

r = 30 0 = × − 420

. gal 7 14. 10 2 s

gal s.

(b) Converting gallons first to liters, then to m 3 ,

r

r

= × F HG^

I KJ

F HG^

I KJ = ×

− −

−

7 14 10 3 786^10

2 3

4

gal s L 1 gal

m 1 L m s

3

3

e j

(c) At that rate, to fill a 1-m 3 tank would take

t = ×

F HG^

I KJ

F HG^

I KJ^

4

m m s

h 3 600

h

3

. 3.^.

Chapter 1 9

P1.29 (a) 6 10 1 1 1 190

F ×^12 HG^

I KJ

F HG^

I KJ

F HG^

I KJ

F HG^

I KJ^

1 000 $ s

h 3 600 s

day 24 h

yr 365 days

years

(b) The circumference of the Earth at the equator is 2 π (^) e 6 378. × 10 3 m (^) j = 4 01. × 107 m. The length of one dollar bill is 0.155 m so that the length of 6 trillion bills is 9 30. × 10 11 m. Thus, the 6 trillion dollars would encircle the Earth

9 30. × (^10 11) 2 32. 104 ×

m= × 4.01 0 m 7 times.

P1.30 N m atoms (^) m Sun atom

kg 1.67 kg

= = × atoms ×

1 99 10 − = ×

30 27

P1.31 V^ = At so t^ = VA = ×^ = ×

3 78 10 −^ −

.^3

m. m

m or m

3 2 b^ μ g

P1.32 V = Bh =

= ×

acres ft acre ft

ft

2

3

a fe j a f

or

V = × ×

F HG^

I KJ = ×

− 9 08 10 2 83^10 1 2 57 10

7 2

6

ft m ft m

3 3 3 3

e j

B

h

B

h

FIG. P1.

P1.33 Fg = (^) b 2 50. tons block (^) ge2 00. × 10 6 blocks (^) jb2 000 lb ton (^) g= 1 00. × 1010 lbs

• P1.34 The area covered by water is

A (^) w = 0 70. A (^) Earth = (^) a0.70 fe 4 π R^2 Earth^ j a= 0.70 fa fe 4 π 6.37 × 10 6 m (^) j^2 = 3 6. × 1014 m^2.

The average depth of the water is

d = (^) a2.3 miles fb1 609 m l mile g = 3 7. × 10 3 m.

The volume of the water is

V = A (^) wd = (^) e3 6. × 10 14 m (^2) je3 7. × 10 3 m (^) j= 1 3. × 1018 m^3

and the mass is

m = ρ V = (^) e1 000 kg m^3 je1 3. × 10 18 m^3 j = 1 3. × 1021 kg.

10 Physics and Measurement

P1.35 (a) d d

d nucleus, scale nucleus, real d

atom, scale atom, real

m ft 1.06 10 m

= ft

F HG^

I KJ^

= ×

×

F HG^

I KJ^ =^ ×

− −

10 e.^ j.^3 , or

d nucleus, scale = (^) e 6 79. × 10 −^3 ft (^) jb304 8. mm 1 ft (^) g= 2 07. mm

(b)

V

V

r r

d d

r r

atom nucleus

atom nucleus

atom nucleus

atom nucleus

m m times as large

= = F HG^

I KJ^

= F HG^

I KJ^

= ×

×

F HG^

I KJ = ×

− −

4 3 4 3

(^3 3 ) 15

3

13

3 3

π π

*P1.36 scale distance between

real distance

scale factor km^

m m

= F km HG^

I KJ

F HG^

I KJ^

= × ×

×

F HG^

I KJ^

− 4 0 10 7 0^10 1 4 10

3

. (^9)

e j.

P1.37 The scale factor used in the “dinner plate” model is

S =

×

0 25 = × −

5.^2

m lightyears

.5 m lightyears.

The distance to Andromeda in the scale model will be

D (^) scale = D (^) actual S = (^) e2.0 × 10 6 lightyears (^) je 2.5 × 10 −^6 m lightyears (^) j = 5 0. m.

P1.38 (a) A A

r r

r r

Earth Moon

Earth Moon^2

Earth Moon

m cm m cm

= = F HG^

I KJ^

×

×

F

H

GG

I

K

(^4) JJ = 4

2 2 6 8

2 π π

e jb. g

(b) (^) VV rr

r r

Earth Moon

Earth Moon

Earth^3 Moon

m cm m cm

F HG^

I KJ^

×

×

F

H

GG

I

K

JJ =

4 3 4 3

6 8

3 3

π π

e jb^ g

P1.39 To balance, m (^) Fe = m Alor ρFe V Fe (^) =ρAl V Al

ρ π ρ π

ρ ρ

Fe Fe Al Al

Al Fe FeAl cm^ cm

3 3

1 3 (^) 1 3

F HG

I KJ^ =^

F HG

I KJ

= F HG^

I KJ^

= FHG IKJ =

r r

r r

/ (^) /

.. .

a f..

12 Physics and Measurement P1.44 A typical raindrop is spherical and might have a radius of about 0.1 inch. Its volume is then approximately 4 × 10 −^3 in 3. Since 1 acre = 43 560 ft 2 , the volume of water required to cover it to a depth of 1 inch is

1 acre 1 inch 1 acre in ft 1 acre

in ft

a fa f a= ⋅ f 6.3 10 6 in^3

F HG^

I KJ

F HG^

I KJ^

43 560 144 ≈ ×

2 2

The number of raindrops required is

n = = × ×

volume of water required − = × volume of a single drop

in in

6 3 3 3

*P1.45 Assume the tub measures 1.3 m by 0.5 m by 0.3 m. One-half of its volume is then

V = a 0 5. fa1 3. mfa 0 5. mfa 0 3. mf =0 10. m 3.

The mass of this volume of water is

m (^) water = ρ (^) water V = (^) e1 000 kg m^3 je 0 10. m^3 j= 100 kg ~ 10 2 kg.

Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The mass of copper required is

m (^) copper = ρ (^) copper V = (^) e8 920 kg m^3 je0 10. m^3 j= 892 kg ~ 10 3 kg.

P1.46 The typical person probably drinks 2 to 3 soft drinks daily. Perhaps half of these were in aluminum cans. Thus, we will estimate 1 aluminum can disposal per person per day. In the U.S. there are ~ million people, and 365 days in a year, so

e^250 ×^10 6 cans day^ jb^365 days year^ g≅^1011 cans

are thrown away or recycled each year. Guessing that each can weighs around 1 10 of an ounce, we estimate this represents

e^10 11 cans^ jb0 1^.^ oz can^ gb^1 lb 16 oz^ gb^1 ton 2 000 lb^ g≈^ 3 1.^ ×^105 tons year.^ ~ 10^5 tons

P1.47 Assume: Total population = 10 7 ; one out of every 100 people has a piano; one tuner can serve about 1 000 pianos (about 4 per day for 250 weekdays, assuming each piano is tuned once per year). Therefore,

tuners ~ 1 tuner

1 000 pianos

1 piano 100 people

people

F HG^

I KJ

F HG^

I KJ^

Chapter 1 13 Section 1.7 Significant Figures

*P1.48 METHOD ONE We treat the best value with its uncertainty as a binomial (^) a 21 3. ± 0 2. (^) f cm (^) a 9 8. ±0 1.fcm ,

A = 21 3 9 8. a. f ± 21 3 0 1. a f. ± 0 2 9 8. a. f a±0 2. fa f0 1. cm 2.

The first term gives the best value of the area. The cross terms add together to give the uncertainty and the fourth term is negligible.

A = 209 cm 2 ± 4 cm^2.

METHOD TWO We add the fractional uncertainties in the data.

A = a 21 3. cm fa9 8. cm f± FHG (^) 21 30 2.. +9 80 1.. IKJ = 209 cm 2 ± 2% = 209 cm^2 ± 4 cm^2

P1.49 (a)^ π^ π π

r^2 2 2 2 2

. m. m m m m m m m

a f (. ) (. )(. ) (. )

(b) 2 π r = 2 πa10 5. m ± 0 2. m f = 66 0. m ±1 3. m

P1.50 (a) 3 (b) 4 (c) 3 (d) 2

P1.51 r m m r

= ± = ± ×

2

43 3

.. cm.. m .. kg

a f a f a f

c h

ρ π

also, δ ρ ρ

= δ^^ m + δ m

r r

In other words, the percentages of uncertainty are cumulative. Therefore,

δ ρ ρ =^ +^ =

a f (^) ,

ρ π

×

1 85− = ×

43 2 3

c h e mj

kg m

and ρ ± δ ρ= a1 61. ± 0 17. f × 10 3 kg m 3 = a1 6. ± 0 2. f× 103 kg m^3.

Chapter 1 15

*P1.57 Consider one cubic meter of gold. Its mass from Table 1.5 is 19 300 kg. One atom of gold has mass

m 0

(^2725) = 197 × 3 27 10

F HG^

I KJ^

= ×

− (^) − u 1.66^10 kg 1 u

a f. kg.

So, the number of atoms in the cube is

N =

×

19 300− 25 kg = 5 90 × 1028 3.27 10 kg

The imagined cubical volume of each atom is

d^312829 5 90 10

×

m (^) = × − m (^3 ) .

So

d = 2 57. × 10 −^10 m.

P1.58 A N A V

V

total drop total A^ V^ r r drop

= = (^) drop 4 total

F H

G

I K

J =

F

H

GG

I

K

a fe j e j (^) π 3 JJe π j 3

A V

total (^) r total m^3 m

= FHG IKJ = × m ×

F HG^

I KJ^

− −

6 5

P1.59 One month is

1 mo = (^) b 30 day (^) gb 24 h day (^) gb3 600 s h (^) g = 2 592. × 10 6 s.

Applying units to the equation,

V = (^) e1 50. Mft 3 mo (^) j t +e 0 008 00. Mft 3 mo^2 j t^2.

Since 1 Mft 3 = 10 6 ft^3 ,

V = (^) e1 50. × 10 6 ft 3 mo (^) j t + (^) e 0 008 00. × 106 ft 3 mo^2 j t^2.

Converting months to seconds,

V = × t t ×

+ ×

×

2

. ft mo. 2 2.592 10 s mo

ft mo 2.592 10 s mo

3 6

3 2

e 6 j.

Thus, V [ft 3 ] = (^) e 0 579. ft 3 s (^) j t + (^) e1 19. × 10 −^9 ft 3 s^2 j t^2.

16 Physics and Measurement

P1.60 (^) α ′(deg) α(rad) (^) tana f α sina f α difference 15.0 0.262 0.268 0.259 3.47% 20.0 0.349 0.364 0.342 6.43% 25.0 0.436 0.466 0.423 10.2% 24.0 0.419 0.445 0.407 9.34% 24.4 0.426 0.454 0.413 9.81% 24.5 0.428 0.456 0.415 9.87% 24.6 0.429 0.458 0.416 9.98% 24.6° 24.7 0.431 0.460 0.418 10.1%

P1.61 2 15 0 2 39

π r r h r h

. m . m tan 55.

a.^ m^ ftan(^.^ )^. m

5555 °°

h

rr

h

FIG. P1.

• P1.62 Let d represent the diameter of the coin and h its thickness. The mass of the gold is

m = V = At = d^ + dh t

F HG^

I KJ

ρ ρ ρ^2 π^ π 4

2

where t is the thickness of the plating.

m = +

L N

M M

O Q

P P

×

= × = =

2

..^...^4

.

. $10 $0..

π^ a^ f^ πa fa f e j

grams cost grams gram cents

This is negligible compared to $4.98.

P1.63 The actual number of seconds in a year is

b86 400 s day gb365.25 day yr g=^ 31 557 600 s yr.

The percent error in the approximation is

π × − × =

(^7) s yr s yr

s yr

e j b^ g^

..

18 Physics and Measurement

P1.68 v =

F HG^

I KJ

F HG^

I KJ

F HG^

I KJ

F HG^

I KJ

F HG^

I KJ

F HG^

I KJ^

5 00 220 = × −

. furlongs.^ 8 32. 10 4 fortnight

yd furlong

m yd

fortnight days

day hrs

hr s

m s

This speed is almost 1 mm/s; so we might guess the creature was a snail, or perhaps a sloth.

P1.69 The volume of the galaxy is

π r t^2 = π e 10 21 m (^) j e^2 10 19 m (^) j~ 1061 m 3.

If the distance between stars is 4 × 10 16 m , then there is one star in a volume on the order of

4 10 16 10

(^3 ) e ×^ m^ j ~^ m^3.

The number of stars is about 10 10

61 50

m 11 m star

stars

3 3 ~^.

P1.70 The density of each material is ρ π π

=^ m = = V

m r h

m (^2) D h 2

Al: g cm cm

g cm

The tabulated value g cm

is smaller.

Cu:

g .23 cm .06 cm

g cm

The tabulated value g cm

is smaller.

Brass: .54 cm .69 cm

g cm

Sn:

g .75 cm .74 cm

g cm

Fe: .89 cm .77 cm

g cm

3 3

3 3

3

3

3

ρ π ρ π ρ π ρ π ρ π = = FHG IKJ

= = FHG IKJ

4 94.4 g 1 5

4 216.1 g 1 9

2

2

2

2

2

b g a f a f b g a f a f b g a f a f b g a f a f b g a f a f

The tabulated value g cm

FHG 7 86. 3 IKJis 0 3%. smaller.

P1.71 (a) (^) b3 600 s hr gb 24 hr day (^) gb365.25 days yr g= 3 16. × 10 7 s yr

(b) V^ r V V

mm cube mm

18

. m. m

m m

1.91 10 micrometeorites

= = × = ×

×

= ×

− −

−

3 7 3 19 3 3 19 3

π π (^) e j

This would take 1 91^10 3 16 10

18 7

×.

×

micrometeorites = × micrometeorites yr

yr.

Chapter 1 19

ANSWERS TO EVEN PROBLEMS

P1.2 5 52. × 10 3 kg m 3 , between the densities of aluminum and iron, and greater than the densities of surface rocks.

P1.34 1 3. × 10 21 kg

P1.36 200 km

P1.4 23.0 kg^ P1.38^ (a) 13.4; (b) 49.

P1.40 r (^) Al r Fe Fe Al

= F HG^

I KJ

ρ ρ

P1.6 7.69 cm^ 1 3

P1.8 (a) and (b) see the solution, N A = 6 022 137. × 10 23 ; (c) 18.0 g; (d) 44.0 g P1.42^ ~ 10^7 rev

P1.10 (a) 9 83. × 10 −^16 g ; (b) 1 06. × 10 7 atoms^ P1.44^ ~ 10^9 raindrops

P1.46 ~ 10 11 cans; ~ 10 5 tons P1.12 (a) 4 02.^ × 10 25 molecules; (b) 3 65. × 10 4 molecules (^) P1.48 a 209 ± 4 f cm^2

P1.14 (a) ii; (b) iii; (c) i (^) P1.50 (a) 3; (b) 4; (c) 3; (d) 2

P1.16 (a) M L T 2

⋅ (^) ; (b) 1 newton = 1 kg m s⋅ 2 P1.52 (a) 797; (b) 1.1; (c) 17.

P1.54 115.9 m P1.18 35 7. m 2 P1.56 316 m P1.20 1 39. × 10 −^4 m 3 P1.58 4 50. m 2 P1.22 (a) 3 39. × 10 5 ft^3 ; (b) 2 54. × 10 4 lb P1.60 see the solution; 24.6° P1.24 (a) 560 km = 5 60. × 10 5 m = 5 60. × 107 cm; (b) 491 m = 0 491. km = 4 91. × 10 4 cm ; P1.62^ 3 64.^ cents^ ; no (c) 6 .19 km = 6 19. × 10 3 m = 6 19. × 105 cm; (^) P1.64 see the solution (d) 2 .50 km = 2 50. × 10 3 m = 2 50. × 105 cm P1.66 (a) 1 000 kg; (b) 5 2. × 10 −^16 kg ; 0 27. kg ; P1.26 4 05 (^10) 1 3. × 10 − (^5) kg

. × 3 m 2

P1.28 (a) 1 mi h = 1 609. km h; (b) 88 5. km h ; (c) 16 1. km h^ P1.68^ 8 32.^ ×^10 −^4 m s^ ; a snail

P1.30 1 19. × 10 57 atoms^ P1.70^ see the solution

P1.32 2 57. × 10 6 m^3